Nonregular Languages Section 2.4 Wed, Oct 5, 2005

Countability of the Set of DFAs Theorem: The set of all DFAs (over an alphabet ) is countable. Proof: For a given n > 0, let Sn be the set of all DFAs with exactly n states. How many DFAs are in Sn? There are n choices for the initial state. For each state, there are n|| choices for the transitions coming out of that state. Therefore, there are (n||)n = n||n choices for .

Countability of the Set of DFAs There are 2n choices for the final states. Therefore, the number of DFAs with exactly n states is n  n||n  2n. The set of all DFAs is S1  S2  S3  … This is a countable set since it is the union of a countable number of finite sets. Thus, we can enumerate the DFAs as M0, M1, M2, M3, …

The Existence of a Non-Regular Language There exists a language that is not accepted by any DFA (provided   ). Proof: Let Ln = L(Mn). Let x be any symbol in . Let sn = xn, for all n  0. Define a new language L by the rule that sn  L if and only if sn  Ln. Then L is not equal to any Ln. So L is not accepted by any DFA.

The Existence of a Non-Regular Language This is another example of a diagonalization argument. It is a non-constructive proof. It does not provide us with an example (unless we actually figure out what each Mn is!).

The Existence of a Non-Regular Language Another non-constructive proof is based on a cardinality argument. The set of all languages is 2*, which is uncountable since its cardinality is equal to the cardinality of 2N, which we know to be uncountably infinite. The set of DFAs is countable. Therefore, the function f(M) = L(M) cannot be onto 2*. So, what is an example of a nonregular language?

The Pumping Lemma The Pumping Lemma: Let L be an infinite regular language. There exists an integer n  1 such that any string w  L, with |w|  n, can be represented as the concatenation xyz such that y is non-empty, |xy|  n, and xyiz  L for every i  0.

Proof of the Pumping Lemma Let n be the number of states. Let w be any string in L with at least n symbols. After processing n symbols, we must have returned to a previously visited state (the Pigeonhole Principle). Let q be the first revisited state. Let x be the string processed from s to q. Let y be the string processed around the loop from q back to q. Let z be the string from q to the end, a final state f.

Proof, continued Then clearly |y| > 0 and |xy|  n. It is also clear that xyiz  L for all i  0, since we may travel the loop as many times as we like, including 0 times.

The Pumping Lemma The Pumping Lemma says that if L is regular, then certain properties hold. The contrapositive of the Pumping Lemma says that if certain properties do not hold, then L is not regular. Therefore, you cannot use the Pumping Lemma to conclude that a language is regular, but only that it is not regular. That’s good, because that is exactly what we want to do.

The Standard Example of a Nonregular Language Let L = {aibi | i  0}. Suppose that L is regular. “Let n be the n of the Pumping Lemma” and consider the string w = anbn. Then w can be decomposed as xyz where |y| > 0 and |xy|  n. Therefore, xy consists only of a’s. It follows that y = ak, for some k > 0.

Standard Example of a Nonregular Language According to the Pumping Lemma, xy2z is in L. However, xy2z =an + kbn, which is not in L. since n + k  n. This is a contradiction. Therefore, L is not regular.

A Second Example of a Nonregular Language Let L = {w  * | w contains an equal number of a’s and b’s}. Suppose L is regular. Let L1 = L(a*b*). Then L  L1 = {aibi | i  0} would also be regular, which is a contradiction. Therefore, L is not regular.

More Examples {w  * | w contains an unequal number of a’s and b’s}. {w  * | w contains more a’s than b’s}. {w  * | w = zz for some z  *}. Consider w = anbanb and use the Pumping Lemma. {w  * | w  zz for some z  *}. Notice that we use the Pumping Lemma only when necessary; other arguments are often simpler.